Optimal. Leaf size=156 \[ \frac{4 a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]
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Rubi [A] time = 0.281975, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3548, 3545, 3544, 205} \[ \frac{4 a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 3548
Rule 3545
Rule 3544
Rule 205
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac{7}{2}}(c+d x)} \, dx &=-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+i \int \frac{(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-(2 a) \int \frac{(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{4 a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\left (4 i a^2\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{4 a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{\left (8 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac{(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{4 a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}\\ \end{align*}
Mathematica [A] time = 3.36801, size = 187, normalized size = 1.2 \[ -\frac{4 i a^2 e^{-i (c+d x)} \cot (c+d x) \left (e^{i (c+d x)} \left (-35 e^{2 i (c+d x)}+26 e^{4 i (c+d x)}+15\right )-15 \left (-1+e^{2 i (c+d x)}\right )^{5/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (-1+e^{4 i (c+d x)}\right )} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.036, size = 412, normalized size = 2.6 \begin{align*}{\frac{{a}^{2}}{15\,d}\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 15\,i\sqrt{ia}\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{3}a+15\,\sqrt{ia}\sqrt{2}\ln \left ({\frac{2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) }{\tan \left ( dx+c \right ) +i}} \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{3}a+76\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia} \left ( \tan \left ( dx+c \right ) \right ) ^{2}+60\,\ln \left ( 1/2\,{\frac{2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a}{\sqrt{ia}}} \right ) \sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{3}a-22\,i\tan \left ( dx+c \right ) \sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}\sqrt{-ia}-6\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia} \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{ia}}}{\frac{1}{\sqrt{-ia}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.81366, size = 1725, normalized size = 11.06 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.42411, size = 1337, normalized size = 8.57 \begin{align*} \frac{\sqrt{2}{\left (208 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 72 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 160 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 120 i \, a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + 15 \, \sqrt{\frac{32 i \, a^{5}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac{{\left (4 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + i \, \sqrt{\frac{32 i \, a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2}}\right ) - 15 \, \sqrt{\frac{32 i \, a^{5}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac{{\left (4 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - i \, \sqrt{\frac{32 i \, a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2}}\right )}{30 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.37264, size = 193, normalized size = 1.24 \begin{align*} \frac{\left (i - 1\right ) \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{4} \log \left (\sqrt{i \, a \tan \left (d x + c\right ) + a}\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} - 6 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a + 14 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} - 16 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + 9 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} - 2 \, a^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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